What do you mean by 'finite calculus'? Also, I presume you mean 'sum over all the integers $k$', not 'where $k$ is any integer' - these are two very different concepts!
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Steven StadnickiSep 6 '11 at 22:57

@Steven Stadnicki . I have edited the question now. Finite calculus is analogous to simple calculus just applied on discrete sets. Here we are working with integers. We might be able to derive the equation using it.
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SanchSep 6 '11 at 23:16

This "trick" is incredibly useful, while so many don't seem to know it.
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TMMSep 7 '11 at 0:17

1

@Thijs: I thought that it was pretty standard: I used to demonstrate it with $\sum_{k=1}^\infty k/2^k$ in my 2nd semester calculus classes, though I’d not care to guess how many really caught on. It is nice, but it’s not finite calculus, so it doesn’t really answer the question.
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Brian M. ScottSep 7 '11 at 0:29

@Steven Stadnicki gave a proper answer. I was thinking that this method is worth mentioning, cos usually I'm using it during my calculus classes.
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Edvin GoeySep 7 '11 at 0:51

@Brian: I agree, but I've seen others (both online and in my studies) struggle with this, when it can be done easily with this trick. And it may not be a proper answer here, but I just think the trick itself is neat. (When seeing discrete sums over discrete objects, I normally would not think of doing "continuous stuff" like taking derivatives and integrals to solve it.)
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TMMSep 7 '11 at 1:17

1

@Arjang, you should be able to undo your downvote now.
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Guess who it is.Sep 7 '11 at 3:55

Different people describe the techniques of finite calculus differently. I like the version presented in Section 2.6 of Graham, Knuth, & Patashnik, Concrete Mathematics, because it shows very clearly the similarity to ordinary calculus. In their notation what you want is $\sum_1^\infty k^2 2^{-k} \delta k$. This is analogous to the ordinary calculus integral $\int_1^\infty x^2 e^{-x} dx$, which you’d do by integrating by parts twice to reduce the $x^2$ factor to a constant. You can do the same thing here using summation by parts. Since this is homework, I’ll do a similar but slightly simpler problem, $\sum_1^\infty k 2^{-k} \delta k$.

@Steven: I like math books with individual character, and it has it in spades: it’s actually fun to read. (Oddly enough, the other two that come to mind are elementary: the liberal arts statistics book by Freedman, Purves, and Pisani, and an old calculus book by Ralph P. Agnew, Calculus: Analytic Geometry and Calculus, with Vectors, which is chock full of snide and sarcastic remarks.)
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Brian M. ScottSep 7 '11 at 9:19

Consider the series
$$
\sum_{k=0}^\infty\;\frac{x^k}{2^k}=\frac{1}{1-x/2}
$$
Take a derivative
$$
\sum_{k=0}^\infty\;k\frac{x^{k-1}}{2^k}=\frac{1/2}{(1-x/2)^2}\tag{1}
$$
Take another derivative
$$
\sum_{k=0}^\infty\;k(k-1)\frac{x^{k-2}}{2^k}=\frac{1/2}{(1-x/2)^3}\tag{2}
$$
Adding $(1)$ and $(2)$ at $x=1$, we get
$$
\sum_{k=0}^\infty\;\frac{k^2}{2^k}=6
$$

You might look at $\sum_{k=1}^n {k^2\over 2^k}$, but this is a fraction, so look at $f(n)=2^n \sum_{k=1}^n {k^2 \over 2^k}$ instead. This satisfies the recurrence $f(n) = 2f(n-1)+n^2$ starting at $f(0)=0$.

It is not difficult to find the values of this for small $n$, namely 0, 1, 6, 21, 58, 141, 318, 685, 1434, 2949, 5998,... and it becomes clear that this is a little less than $6\times 2^n$ much as you might expect from the question. It is not difficult then to take differences (or looking up OEIS A047520, which I considered 11 years ago) and suspect it is likely to satisfy $f(n) = 6\times 2^n -n^2-4n-6$, and this can easily be proved by induction.

For starters, you know that $\displaystyle \sum_{k=1}^\infty {1\over 2^k} = 1$, yes? If you don't have that much, then you likely won't be able to follow the rest of the proof.

Now, consider $\displaystyle S=\sum_{k=1}^\infty {k\over 2^k}$. Then $\displaystyle {S\over 2} = \sum_{k=1}^\infty {k\over 2*2^k} = \sum_{k=1}^\infty {k\over 2^{k+1}} = \sum_{l=2}^\infty {l-1\over 2^l} = \sum_{l=1}^\infty {l-1\over 2^l}$. (The reason why the last equality is true is because the $l=1$ term is ${1-1\over 2^1}=0$, so we can add it in without changing the sum.) This means that $\displaystyle S-{S\over 2} = \sum_{k=1}^\infty {k\over 2^k} - \sum_{k=1}^\infty {k-1\over 2^k} = \sum_{k=1}^\infty {k-(k-1)\over 2^k} = \sum_{k=1}^\infty {1\over 2^k} = 1$, or in other words $S=2$. Now, can you see how to use a similar technique on your sum? The terms will be a little more complicated, but using the formula for $\sum_{k=1}^\infty {k\over 2^k}$ will help.